Question

Rima was given two test tubes labelled A and B. When she dipped an iron nail in both the test tubes, she observed that the colour of the solution changed as shown in the image given below. Identify the solutions in test tubes A and B.

• A. Solution in test tube A is ${\text{FeSO}}_4$ and that in test tube B is ${\text{CuSO}}_4$
• B. Solution in test tube A is ${\text{ZnSO}}_4$ and that in test tube B is ${\text{CuSO}}_4$
• C. Solution in test tube A is ${\text{CuSO}}_4$ and that in test tube B is ${\text{ZnSO}}_4$
• D. Solution in test tube A is ${\text{CuSO}}_4$ and that in test tube B is ${\text{FeSO}}_4$

Answer 1

The correct option is D Solution in test tube A is ${\text{CuSO}}_4$ and that in test tube B is ${\text{FeSO}}_4$

The blue coloured solution in test tube A is ${\text{CuSO}}_4$ while the green coloured solution in test tube B is ${\text{FeSO}}_4$. When an iron nail is dipped in test tube A, $\text{Fe}$ displaces $\text{Cu}$ from ${\text{CuSO}}_4$ solution as $\text{Fe}$ is more reactive than $\text{Cu}$. This causes the solution to turn green. The reaction occurs as follows:
$Fe\left(s\right)+CuS{O}_4\left(aq\right)\to FeS{O}_4\left(aq\right)+Cu\left(s\right)$

However, when an iron nail is dipped in test tube B containing ${\text{FeSO}}_4$ solution, the colour of the solution remains the same. This is because, $\text{Fe}$ of the nail cannot displace $\text{Fe}$ of ${\text{FeSO}}_4$ solution.